Question #b00d1

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Answer 1 · 2015-11-03T17:50:25

$abs(b+a) = sqrt(3)$

Consider the diagram below.
$vec(b)=vec(PQ)$ and $abs(vec(PQ))=abs(vecb)=1$

$veca = vec(PQ)=vec(TS)$ and $abs(vec(PQ)) = abs(vec(TS)) = abs(veca)=1$

$vec(a+b) = vec(PS)$

$/_STU = /_QPT = 60^@$ (the angle between $vec(a)$ and $vec(b)$
$rArr /_PTS = 120^@$
$rArr /_RPT =/_RST=30^@$

The bisector of $/_PTS$
divides $triangle PTS$ into 2 congruent triangles
$abs(PR) = abs(RS)$

Since $/_PTR=60^@$ and $abs(PT)=1$
$rArr abs(PR) = sqrt(3)/2$

and
$abs(PS) = sqrt(3)$

Therefore $abs(vec(a+b)) = sqrt(3)$

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